Question

A thin sheet of metal of uniform thickness is cut into the shape bounded by the line $x=a$ and $y=\pm k{x}^2$ as shown.
Find the coordinates of the centre of mass.
[Here, $a\&k$ are positive constant]

• A. $(\frac{3a}{2},0)$
• B. $(\frac{3a}{4},0)$
• C. $(0,\frac{3a}{4})$
• D. $(3a,0)$

Answer 1

The correct option is B $(\frac{3a}{4},0)$


Consider an element of thickness $dx$ at distance $x$ as shown in the figure.

The mass of this element is given by

$dm=\sigma \times dx\times 2y=2\sigma ydx$

Where, $\sigma$ is the areal density of metal sheet.

The $x-$coordinate of the centre of mass is given by

$x_{com}=\frac{\int xdm}{\int dm}=\frac{\underset{0}{\overset{a}{\int }}x\times 2\sigma ydx}{\underset{0}{\overset{a}{\int }}2\sigma ydx}$

$x_{com}=\frac{\underset{0}{\overset{a}{\int }}x\times k{x}^{2}dx}{\underset{0}{\overset{a}{\int }}k{x}^{2}dx}[\because y=k{x}^2]$

$\Rightarrow x_{com}=\frac{\underset{0}{\overset{a}{\int }}{x}^{3}dx}{\underset{0}{\overset{a}{\int }}{x}^{2}dx}=\frac{\frac{a^4}{4}}{\frac{a^3}{3}}$

$\therefore x_{com}=\frac{3a}{4}$

Since the sheet is symmetrical about the $x-$axis, we can say that $y_{com}=0$.

$\therefore$ coordinates of centre of mass are $(\frac{3a}{4},0)$

Hence, option $\left(B\right)$ is the correct answer.